# Exercise--Related rates and implicit differentiation

1. Each edge of a cube is expanding at the rate of 1 centimeter (cm) per second. How fast is the volume changing when the length of each edge is (a) 5 cm? (b) 10 cm? (c) cm?

2. An airplane flies in level flight at constant velocity, eight miles above the ground. (In this exercise assume the earth is flat.) The flight path passes directly over a point P on the ground. The distance from the plane to P is decreasing at the rate of 4 miles per minute at the instant when this distance is 10 miles. Compute the velocity of the plane in miles per hour.

3. A baseball diamond is a 90-foot square. A ball is batted along the third-base line at a constant speed of 100 feet per second. How fast is its distance from first base changing when (a) it is halfway to third base? (b) it reaches third base?

4. A boat sails parallel to a straight beach at a constant speed of miles per hour, staying miles offshore. How fast is it approaching a lighthouse on the shoreline at the instant it is exactly miles from the lighthouse ?

5. A reservoir has the shape of a right-circular cone. The altitude is 10 feet, and the radius of the base is 4 ft. Water is poured into the reservoir at a constant rate of 5 cubic feet per minute. How fast is the water level rising when the depth of the water is 5 feet if (a) the vertex of the cone is up? (b) the vertex of the cone is down?

6. A water tank has the shape of a right-circular cone with its vertex down. Its altitude is 10 feet and the radius of the base is 15 feet. Water leaks out of the bottom at a constant rate of 1 cubic foot per second. Water is poured into the tank at a constant rate of cubic feet per second. Compute so that the water level will be rising at the rate of 4 feet per second at the instant when the water is 2 feet deep.

7. Water flows into a hemispherical tank of radius feet (flat side up). At any instant, let denote the depth of the water, measured from the bottom, the radius of the surface of the water, and the volume of the water in the tank. Compute at the instant when feet. If the water flows in at a constant rate of cubic feet per second, compute , the rate at which is changing, at the instant when feet.

8. A variable right triangle in the -plane has its right angle at vertex , a fixed vertex at the origin, and the third vertex restricted to lie on the parabola The point starts at the point at time and moves upward along the -axis at a constant velocity of cm/sec. How fast is the area of the triangle increasing when sec?

9. The radius of a right-circular cylinder increases at a constant rate. Its altitude is a linear function of the radius and increases three times as fast as the radius. When the radius is 1 foot the altitude is 6 feet. When the radius is 6 feet, the volume is increasing at a rate of 1 cubic foot per second. When the radius is 36 feet, the volume is increasing at a rate of cubic feet per second, where n is an integer. Compute .

10. A particle is constrained to move along a parabola whose equation is . (a) At what point on the curve are the abscissa and the ordinate changing at the same rate? (b) Find this rate if the motion is such that at time we have and .

11. The equation defines as one or more functions of . (a) Assuming the derivative exists, and without attempting to solve for , show that satisfies the equation . (b) Assuming the second derivative exists, show that whenever .

12. If , the equation defines as a function of . Without solving for , show that the derivative has a fixed sign. (You may assume the existence of .)

13. The equation defines implicitly as two functions of if . Assuming the second derivative exists, show that it satisfies the equation .

14. The equation defines implicitly as a function of . Assuming the derivative exists, show that it satisfies the equation

15. If , where is a rational number, say , then . Assuming the existence of the derivative , derive the formula using implicit differentiation and the corresponding formula for integer exponents.

# L'Hôpital's Rule

The Cauchy Mean Value Theorem is the basic tool needed to prove a theorem which facilitates evaluation of limits of the form

when

Theorem. Suppose that

and and suppose also that

exists. Then

exists, and

PROOF. The hypothesis that exists contains two implicit assumptions:

(1) there is an interval such that and exist for all in except, perhaps, for ,

(2) in this interval with, once again, the possible exception of

On the other hand, and are not even assumed to be defined at . If we define (changing the previous values of and , if necessary), then and are continuous at . If , then the Mean Value Theorem and the Cauchy Mean Value Theorem apply to and on the interval (and a similar statement holds for ). First applying the Mean Value Theorem to , we see that , for if there would be some in with , contradicting (2). Now applying the Cauchy Mean Value Theorem to and , we see that there is a number . in such that

or

Now approaches as approaches , because , is in ; since exists, it follows that

# Uniform Continuity

Definition. The function is uniformly continuous on an interval if for every there is some such that, for all and in

Lemma. Let and let be continuous on the interval Let , and suppose that statements (i) and (ii) hold:

(i) if and are in and , then ,

(ii) if and are in and , then

Then there is a 0 such that

Proof. Since is continuous at , there is a 0 such that

It follows that

(iii) if and then

Choose to be the minimum of and . We claim that this works. In fact, suppose that and are any two points in with . If and are both in then by (i); and if and are both in then by (ii). The only other possibility is that

or

In either case, since , we also have and So by (iii).

THEOREM If is continuous on , then is uniformly continuous on

PROOF: For let's say that is good on if there is some such that, for all and in if , then . Then we're trying to prove that is -good on for all

Consider any particular . Let

Then (since a is in ), and is bounded above (by ), so has a least upper bound . We really should write , since and might depend on . But we won't since we intend to prove that , no matter what is.

Suppose that we had . Since is continuous at , there is some such that, if , then . Consequently, if and , then So is surely -good on the interval . On the other hand, since is the least upper bound of , it is also clear that is -good on . Then the above Lemma implies that is -good on , so is in A, contradicting the fact that is an upper bound.

To complete the proof we just have to show that is actually in . The argument for this is practically the same: Since is continuous at , there is some such that, if , then So is -good on . But is also -good on , so the Lemma implies that is -good on .

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